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Hello all,
I'm learning C and I am doing a code review. Now I am looking at a line I have no clue what it's doing... could you guys help me out? It says:
int number = (array[3] - '0') - 1; // array is of the type char * and in the position 3 will have a 1,2,3 or 4.
The part I don't get is the "- '0'" part...
Thank you!
That's a common syntax to remove the offset for digits, from ascii values.
Using that, the char '4', becomes the digit 4. It works for all digits because they are all consecutively valued by the ascii numbering system.
If you don't have a copy of the ascii table, you should google it, and grab one. You'll find it very useful indeed.
C uses ASCII character values:
ASCII Table / Extended ASCII Codes
Single quoted characters stand in for their ascii value, eg
is true.Code:if ('0' == 48)
You can see the ascii values this way:
As Adak points out, using ascii values are a surprisingly fundamental aspect of C programming.Code:char string[]="hello world";
int i;
for (i=0;i<11;i++) printf("%d ",string[i]);
'0' will take the ASCII value of zero.Code:main()
{
char arr[]={1,2,3,4};
int number=(arr[3]-'0')-1;
printf("%d\n",number);
}
ASCII value of zero is 48.
In the above code the value of arr[3] is 4.
So (4-48)-1 is equal to (-44)-1=>-45.
It is Just default type conversion. Normally character values will convert into integer based on its ASCII value.
For Ex: '0' will be taken as 48 in ASCII value. So C follows this ASCII standard for conversion.
To be perfectly pedantic, C doesn't require ASCII. It does, however, require that '0', '1', .. '9' be consecutive, so the idiom will always work.
No it doesn't. Your implementation does. The standard says nothing about '0' = 48.Quote:
Originally Posted by sganesh
No again. Your implementation does...Quote:
Originally Posted by MK27
thanks guys!